\(\int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\) [230]

Optimal result

Integrand size = 28, antiderivative size = 162 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

(1/8-1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2 
))/a^(5/2)/d+1/5*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(5/2)+13/30*tan(d*x 
+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)+67/60*tan(d*x+c)^(1/2)/a^2/d/(a+I*a 
*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^2(c+d x) \sqrt {\tan (c+d x)} \left (-2 (19+86 \cos (2 (c+d x))+80 i \sin (2 (c+d x)))-\frac {15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{\sqrt {i a \tan (c+d x)}}\right )}{120 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[1/(Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]
 

Output:

(Sec[c + d*x]^2*Sqrt[Tan[c + d*x]]*(-2*(19 + 86*Cos[2*(c + d*x)] + (80*I)* 
Sin[2*(c + d*x)]) - (15*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/S 
qrt[a + I*a*Tan[c + d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[a 
+ I*a*Tan[c + d*x]])/Sqrt[I*a*Tan[c + d*x]]))/(120*a^2*d*(-I + Tan[c + d*x 
])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4027, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {9 a-4 i a \tan (c+d x)}{2 \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {9 a-4 i a \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {9 a-4 i a \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {41 a^2-26 i a^2 \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {41 a^2-26 i a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {41 a^2-26 i a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {67 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {15}{2} a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {67 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {15}{2} a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {67 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\frac {\frac {67 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\left (\frac {15}{2}-\frac {15 i}{2}\right ) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {67 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {13 a \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

Int[1/(Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]
 

Output:

Sqrt[Tan[c + d*x]]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((13*a*Sqrt[Tan[c 
+ d*x]])/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((15/2 - (15*I)/2)*a^(3/2)* 
ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/ 
d + (67*a^2*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(6*a^2))/( 
10*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] 
 && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 575 vs. \(2 (128 ) = 256\).

Time = 1.64 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.56

method	result	size
derivativedivides	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}-90 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+268 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{3}+60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -1060 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+908 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}-420 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{240 d \,a^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{4} \sqrt {-i a}}\)	\(576\)
default	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}-90 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+268 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{3}+60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -1060 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+908 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}-420 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{240 d \,a^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{4} \sqrt {-i a}}\)	\(576\)

Input:

int(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(15*I*2^(1/2)*ln(-(-2 
*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d* 
x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^4-90*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^( 
1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c) 
+I))*a*tan(d*x+c)^2+268*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 
2)*tan(d*x+c)^3+60*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I 
*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3+15* 
I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 
2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-1060*I*(-I*a)^(1/2)*(a*tan(d*x+c) 
*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-60*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2 
)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I) 
)*a*tan(d*x+c)+908*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan( 
d*x+c)^2-420*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a^3/(a*ta 
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (120) = 240\).

Time = 0.10 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.04 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (30 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 30 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (83 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 102 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 22 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas 
")
 

Output:

1/120*(30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(a^3*d*sqrt( 
-1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c 
) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2 
*I*d*x + 2*I*c) + 1)) - 30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c 
)*log(-a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/( 
e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 
2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I 
*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(83 
*e^(6*I*d*x + 6*I*c) + 102*e^(4*I*d*x + 4*I*c) + 22*e^(2*I*d*x + 2*I*c) + 
3))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {\tan {\left (c + d x \right )}}}\, dx \] Input:

integrate(1/tan(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral(1/((I*a*(tan(c + d*x) - I))**(5/2)*sqrt(tan(c + d*x))), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima 
")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(1/(tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)
 

Output:

int(1/(tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {1}{\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

int(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(1/(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2 
*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + 
 d*x))*sqrt(tan(c + d*x)*i + 1)),x))/(sqrt(a)*a**2)